Define the following residue for using the residue theorem I wanna calculate the following integral
$$\oint_{\{\vert z \vert =\frac{5}{2}\}} \frac{1}{sin^2(z)cos(z)}dz$$
with the residuum theorem.
Well first I have to calculate the residue of this integral. We have that $cos(z)=0 \iff \frac{\pi}{2}+k \pi, k \in \Bbb Z$ and $sin(z)=0 \iff z=k \pi, k \in \Bbb Z$. This means, we get the following isolated zero:

*

*$z_0=0$

*$z_1=\frac{\pi}{2}$

*$z_2=-\frac{\pi}{2}$
Now I have to calculate the residue of the different $z_i$ but for this I have to know, which kind of isolated zero we have. Here is where I got stucked.. How can I find this out?
EDIT: For the real case we have that $sin^2(x)cos(x)=cos(x)-cos^3(x)$. Can I maybe use it here?
 A: To find out which kind of isolated zero you have, you can have a look on the series expansion of $\sin$ and $\cos$ in the point $z_i$. For $z_0=0$ the terms of lowest order of $\sin$ is $z$ and of $\cos$ it is $1$. This means the term of lowest order of $\sin^2(z)\cos(z)$ is $z^2$ and thus you have a second-order pole. Then you can do the same for $z_1$ and $z_2$, where you will find out that you have simple poles.
Another way to do so is you try to find out for which $n$
$$\frac{(z-z_i)^n}{\sin^2(z)\cos(z)}
$$
is holomorphic and nonzero in a neighbourhood of $z_i$, what is actually the defintion of the order of a pole.
A third way is to calculate the laurent series, which is a very similar method to the first method, but delivers you directly your residue. Again for example for the point $z_0=0$, you have
$$ \sin z=z+\mathcal O(z^3)
$$and
$$ \cos z=1+\mathcal O(z^2)
$$
thus
\begin{align}\frac{1}{\sin^2(z)\cos(z)}&=\frac{1}{(z+\mathcal O(z^3))^2(1+\mathcal O(z^2))}=\frac{1}{z^2(1+\mathcal O(z^2))^2(1+\mathcal O(z^2))}\\
&=\frac{1}{z^2(1+\mathcal O(z^2))(1+\mathcal O(z^2))}=\frac{1}{z^2(1+\mathcal O(z^2))}\\
&=\frac{1}{z^2}(1+\mathcal O(z^2)).
\end{align}
In the last step we have used the geometric series. So your residue in $z_0=0$ is $0$.
